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11 Jun 2019, 01:41
Kudos
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D
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Difficulty:
25%
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Question Stats:
78% (01:52) correct
22%
(02:07)
wrong
based on 193
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
11 Jun 2019, 02:27
Kudos
Bookmarks
Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
Prob=1-All three are green; 1-(2/5)^3=117/125 IMO D
11 Jun 2019, 05:51
Kudos
Bookmarks
Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
A. \(\frac{8}{125}\)
B. \(\frac{36}{125}\)
C. \(\frac{108}{125}\)
D. \(\frac{117}{125}\)
E. \(\frac{124}{125}\)
Probability not all disks are green = 1 - Probability that all disks chosen are green
--> Prob that all disks chosen are green = 2/5*2/5*2/5 = 8/125
--> Required Probability = 1 - 8/125 = 117/125
IMO Option D
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